The trace property used in this derivation is key to simplifying certain matrix expressions. Here's a summary of how the trace property works in the context of the image:

Trace Property of a Matrix:

The trace of a matrix $\text{tr}(A)$ is the sum of the diagonal elements of the matrix $A$.

1. Linearity of Trace: $\text{tr}(A + B) = \text{tr}(A) + \text{tr}(B)$ $\text{tr}(cA) = c \cdot \text{tr}(A)$ where $A$ and $B$ are matrices, and $c$ is a scalar.

2. Cyclic Property of the Trace: For matrices $A$ and $B$ of compatible dimensions: $\text{tr}(AB) = \text{tr}(BA)$ This allows reordering of matrices inside the trace, which is useful for simplification.

Application in the Image:

In Step 5 of the derivation:

• The term $\text{tr}(AV)$ appears, where $V$ is the covariance matrix of the deviation $(y - \mu)$.
• The cyclic property of the trace is likely used to reorder the terms inside the trace, enabling simplification.

In the final step:

$E[y'Ay] = \mu'A\mu + \text{tr}(AV)$ This result uses the fact that the trace $\text{tr}(AV)$ simplifies the expected value of the quadratic form involving $A$ and the deviation term $(y - \mu)$.

Do you want any further clarification or details?

Here are 5 related questions you might find useful:

1. What is the covariance matrix $V$ and how is it related to $(y - \mu)$?
2. How does the expectation operator apply to matrix equations like these?
3. What are the properties of quadratic forms in matrix algebra?
4. Why does $\mu'A(y - \mu)$ equal zero in this derivation?
5. How is the trace of a matrix computed in practical terms?

Tip: Remember that the cyclic property of the trace is very useful in matrix algebra because it allows you to rearrange terms, making some computations more straightforward.